Question

D 4.37. Assuming the availability of diodes for which vD = 0.75 V at iD = 1mA, design a circuit that utilizes four diodes connected in series, in series with a resistor R connected to a 15-V power supply. The voltage across the string of diodes is to be 3.3 V.

Answer 1

Answer:

R = 0.5825 k ohm

Explanation:

given data

vD = 0.75 V

iD = 1mA

resistor R = 15-V

solution

we get here first vD that is

vD = $\frac{3.3}{4}$  

vD = 0.825

so

iD = Ise × $e^{\frac{vD}{nvT}}$      

n = 1

so we can say

$\frac{iD2}{iD1} = e^{\frac{vD2-vD1}{nvT}}$  

so it will

iD2 = iD1 × $e^{\frac{vD2-vD1}{nvT}}$

put here value and we get

iD2 = 1 × $e^{\frac{0.825-0.75}{25}}$  

iD2 = 1 × $e^{3}$  

iD2 = 20.086 mA

so

R will be

R = $\frac{15-3.3}{ID}$  

R = 0.5825 k ohm